Question

Limit of Riemann sum problem

Answer 1

\[\int\limits_{0}^{1}(x^2 + x)dx\]

Answer 2

I'm having a hard time following how to do this with my book's example. Can someone explain the steps to solve it?

Answer 3

Are you ready for some algebra???? That's all it is.

Right Side Only:

One Interval: h = 1, 1(1+1) = 1(2) = 2

Two Intervals: h = 1/2, (1/2)(1/4 + 1/2) + (1/2)(1+1)= (1/2)(3/4) + (1/2)(2) = 3/8 + 1 = 11/8 = 1.375

Four Intervals: h = 1/4

(1/4)[(h^2 + h) + ((2h)^2 + 2h) + ((3h)^2 + 3h) + ((4h)^2 + 4h)] =
(1/4)[(h^2)(1+4+9+16)+h(1+2+3+4)] = 35/32 = 1.094

Let's think about that for a moment...

Eight Intervals: h = 1/8

(h)[(h^2 + h) + ((2h)^2 + 2h) + ((3h)^2 + 3h) + ... + ((8h)^2 + 8h)] =
(h)[(h^2)(1+4+9+...+64)+h(1+2+3+...+8)] = 123/128 = 0.961

Our task, then, is to add up two things:

Some number of Whole Numbers: 1 + 2 + 3 + ... + n
The same number of their squares: 1^2 + 2^2 + 3^2 + ... + n^2

Conveniently, these are known sums.

Some number of Whole Numbers: 1 + 2 + 3 + ... + n = n(n+1)/2
The same number of their squares: 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6

m intervals: h = 1/m

(1/m)[((1/m)^2)(m(m+1)(2m+1)/6)+(1/m)(m(m+1)/2)] =
[((1/m)^2)((m+1)(2m+1)/6)+(1/m)((m+1)/2)] =

\(\dfrac{(m+1)(5m+1)}{6m^{2}}\) = \(\dfrac{1}{m} + \dfrac{1}{6m^{2}} + \dfrac{5}{6}\)

This last expression clearly approaches 5/6 as m increases without bound.

Wasn't that fun?! Everyone should do this a couple of times. We're trying to reinvent the calculus. It is kind of amazing how much algebra we needed to do that!

Answer 4

Yup I have learned this. But I have to do this problem like this:
And then I believe

Is it like this?